CHEMISTRY COMMUNITY

Tiffany Chen 1A

Sigma and pi bonds don't really correlate with s and p, respectively. Sigma bonds form when an s orbital overlaps with another s orbital, an s overlaps with a p, or a p overlaps with another p end-to-end. Pi bonds are formed when additional orbitals overlap side-by-side.

Tiffany Chen 1A

I think in general, since strong acids and bases are completely ionized in water, the more soluble an acid/base is in water, the stronger it is. So for this question, BrO- would be the stronger base.

Tiffany Chen 1A

If you're asking about how to tell which part of the ligand binds to the metal, the atom (which is part of the ligand) with a lone pair is the one that's directly bonded to the central metal atom.

Tiffany Chen 1A

Since the 2 lone pairs of H2O are on the same O atom, H2O can't bind to 2 different sites on a central metal atom. Therefore, H2O can bind to a maximum of only 1 site on a central metal atom, making it monodentate.

Tiffany Chen 1A

You can also think about it in terms of resonance. The first structure would have resonance (you can move the double bonds around and get the same structure), which means the double bonds are actually "spread out" throughout all the bonds. So, the actual (observed) bond of the circled bond...

Tiffany Chen 1A

(In the context of the test question) A resonance structure is like a variation to the given structure, but the resonance structure has to be equivalent to the given structure. So, if there's 1 double bond in a structure and the rest of the bonds are single bonds, you can't just move the double bond...

Tiffany Chen 1A

I don't think you have a lone pair on the middle N and have all 16 electrons accounted for and all the octets. If you put a lone pair in the middle, then either oxygen won't have its octet or the other nitrogen won't have its octet.

Tiffany Chen 1A

The extra electron isn't really designated to a specific atom. You just have to account for the extra electron when you're adding up all the valence electrons in the molecule and then draw the structure the same way you would draw any other structure. Also, just put brackets around the structure wit...

Tiffany Chen 1A

To calculate the formal charges of atoms in a molecule, you use the typical number of valence electrons for that atom. But if you added the formal charges of all the atoms in an ion, they should add up to the total charge of the ion. For example, in CH3(-), C has a formal charge of -1 (FC=4-2-3=-1) ...

Tiffany Chen 1A

I drew one structure that had a double bond between the carbon and oxygen (all other bonds are single bonds) and then the other structure had a double bond between the nitrogen and carbon. The first structure would have 1 lone pair on N and the second structure would have a single bond between O and...

Tiffany Chen 1A

Oh wait is it because N wouldn't have its octet?

Tiffany Chen 1A

For part A, the question asks for the Lewis structure and formal charges of NO+. The back of the book says the correct Lewis structure is N and O attached by a triple bond, where N has 1 lone pair of e- and O has 1 lone pair of e-. So the formal charges would be 0 for N and +1 for O. I understand ho...

Tiffany Chen 1A

(Just wanted to clarify) When you draw the lewis structure for an ionic bond between a cation and an anion, do you just write the element of the cation with a positive charge next to the element of the anion with a negative charge? And you would draw out the electrons for only the anion but not the ...

Tiffany Chen 1A

If the total # of electrons is an odd number, then there is a radical.

Tiffany Chen 1A

I think the main group metals are the metals in the s block and p block (metals in groups 1, 2, 13, 14, 15, 16, 17, and 18). For elements like Ga, In, Sn, Tl and Pb, I don't think they have to lose all of their electrons in their d orbital, because their valence electrons reside in only the s and p ...

Tiffany Chen 1A

Electron affinity is how likely an atom will take an electron or how much it "wants" an electron. Electronegativity is how strongly an atom will attract shared electrons to itself (more related to bonding). Both electron affinity and electronegativity show similar trends on the periodic t...

Tiffany Chen 1A

It's because this isn't always true, kinetic energy also depends on what the threshold energy or work function is.

Tiffany Chen 1A

Electron affinity is how likely an atom will take an electron or how much it "wants" an electron. Electronegativity is how strongly an atom will attract shared electrons to itself (more related to bonding). Both electron affinity and electronegativity show similar trends on the periodic ta...

Tiffany Chen 1A

Oh thanks for letting me know, I didn't know that! So I guess tungsten is the exception to the exception?
From high school chem I know that the rules get more complicated as you move down the periodic table, but I wouldn't worry too much about it since we should really just focus on the first 4 rows.

Tiffany Chen 1A

If l=1, this means that the possible ml values are 1, 0, and -1. Since there are 3 possible ml values, there are 3 different orbital states in which electrons can exist. Since 1 orbital can hold up to 2 electrons, 3 orbitals can hold up to 6 electrons. Another way to think about it is by looking at ...

Tiffany Chen 1A

You shouldn't have to memorize any electron configurations, you should just know how to figure out the electron configurations of an element when asked. So you should know how to read the periodic table (s-, p-, and d-blocks) and be able to write down the corresponding e- configuration using the rul...

Tiffany Chen 1A

I figured this problem out by first solving for n2 when the wavelength is 656.3 nm (first line in the Balmer series). It's a fact that the Balmer series starts at n=2, so you know that n1=2. Then using hc/lambda = -(hR/n^2)+(hr/n^2) where lambda=656.3, the first n=2, and the second n is 3. So now yo...

Tiffany Chen 1A

You could use De Broglie's equation: lambda=h/mV, where h is Planck's constant, m is the mass of the electron (which is a constant that would be given to you or you could just look it up) in kg, and V is the velocity in m/s.

Tiffany Chen 1A

I think you can just use E=hv to solve for the threshold energy. So E=(6.626*10^-34)(2.50*10^16) should give you the correct answer.

Tiffany Chen 1A

There's a kg "hidden" in h. The units for h are J*s and a Joule is the same as kg*(m^2)*(s^-2). Taking this into account, the kg from h will cancel with the kg from mass, and the other units also cancel until you're left with meters.

Tiffany Chen 1A

So kinetic energy is this case would be \frac{1}{2}mV^{2} and we would just plug in the mass of an electron and the given velocity? Also, how would you be able to tell whether or not you should account for kinetic energy? Do we include kinetic E here, because we know that this problem is about the p...

Tiffany Chen 1A

Since you're given the velocity of an electron, to do part A, you can use the De Broglie equation lambda = h/mV where m is the mass of an election (you can just look this up, it's 9.11x10^-31) and V is the velocity and then solve for the wavelength.

Tiffany Chen 1A

32 W is the same as 32 J/s. Since you want to find the amount of photons generated after 2 seconds, multiply 32 J/s by 2 s (seconds cancel out) and then you're left with 64 J. 64 J is the total energy emitted (not sure if emitted is the right word) in 2 seconds. After using E=hc/lambda to find the e...

Tiffany Chen 1A

Yup! The molecular formula tells you the actual number of atoms for each element, so you could just count them all up when given a diagram of a molecule. Then, like you said, you can simply the molecular formula to get the empirical formula (unless you can't simplify, then the molecular and empirica...

Tiffany Chen 1A

E=hv calculates the amount of energy per photon of a light source with that particular frequency, so all photons of that light source would have the same amount of energy. I think the "low frequency light has no photons with sufficient energy" from lecture is referring to the 1 to 1 photon...

Tiffany Chen 1A

Yeah, your answer 0.000000522 (or 522 x 10^-9) is in meters and you could just leave it as meters unless the problem specifically asks for a wavelength in nanometers (or any other specific unit). If you want to convert your answer from m to nm, you can move the decimal 9 places to the right or multi...

Tiffany Chen 1A

To convert from meters to nanometers, we just multiply the number of meters by 10^9 right? And then to convert meters to picometers, we just multiply by 10^12?

Tiffany Chen 1A

Thanks!

Tiffany Chen 1A

When finding the empirical formula, you're ultimate goal is to have all the numbers as integers (no decimals), but when solving for the ratio you usually end up with decimals and it's hard to get rid of all of the decimals in one step. By dividing all of the numbers by the smallest number, there's a...

Tiffany Chen 1A

I solved for the molecular formula differently than the other students that replied in the other post about this homework question and I just wanted to make sure that my method of solving was ok. Instead of solving for the empirical formula first and then solving for the molecular formula, I just so...

Tiffany Chen 1A

In this particular problem, I think it's kind of confusing to think about the ratio of CaC2 to C2H2 as 1:2 instead of 1:1 because that would make the equation unbalanced. But if you had a different chemical equation where the ratio of the limiting reactant to the product the question referred to was...

CHEMISTRY COMMUNITY

Search found 36 matches

Re: pi bond and sigma bond

Re: 12.49 Determining Base Strength

Re: Ligand binding

Re: Problem 17.33

Re: Test 3 Q 8 [ENDORSED]

Re: Resonance Review

Re: 4.13 D

Re: Ions

Re: Formal charge with ions

Re: Worksheet 7 #2

Re: 3.49

3.49

Ionic Lewis Structures

Re: Radicals [ENDORSED]

Re: Main group metals

Re: question

Re: Kinetic Energy

Re: question

Re: Electron Configuration Exceptions

Re: Midterm Review Q5 part f [ENDORSED]

Re: Electron Configuration Exceptions

Re: Knowing the Balmer series HW q1.57

Re: Calculating wavelength of an electron

Re: Question 1.33, part b

Re: 1.33 [ENDORSED]

Re: Homework 1.33 Part C

Re: HW 1.33

Re: HW 1.27

Re: Empirical and Molecular Formulas [ENDORSED]

Re: Photons in a light source [ENDORSED]

Re: Units for Wavelengths [ENDORSED]

Clarification on unit conversions

Re: Alternative way to solve for F.19

Re: Why Divide by Smallest Number? [ENDORSED]

Alternative way to solve for F.19

Re: In-Class Limiting Reactant and Theoretical Yield Problem [ENDORSED]